a) Ta có: \(2\left|x-4\right|+\left|2x+5\right|=12\)
\(\Leftrightarrow4\cdot\left(x-4\right)+2x+5+4\left|x-4\right|\left|2x+5\right|=144\)
\(\Leftrightarrow4x-16+2x+5-144=-4\left|x-4\right|\left|2x+5\right|\)
\(\Leftrightarrow6x-155=-4\left|x-4\right|\left|2x+5\right|\)
\(\Leftrightarrow36x^2-12x+24025=16\left(x-4\right)\left(2x+5\right)\)
\(\Leftrightarrow36x^2-12x+24025=16\left(2x^2+5x-8x-20\right)\)
\(\Leftrightarrow36x^2-12x+24025-16\left(2x^2-3x-20\right)=0\)
\(\Leftrightarrow36x^2-12x+24025-32x^2+48x+320=0\)
\(\Leftrightarrow4x^2+36x+24345=0\)
\(\Leftrightarrow\left(2x\right)^2+2\cdot2x\cdot9+81+24264=0\)
\(\Leftrightarrow\left(2x+9\right)^2+24264=0\)(Vô lý)
Vậy: \(x\in\varnothing\)
